I'm trying to change a table field that contains decimal numbers from varchar(255) to decimal(12,2). And before I do that, I'd like to find out if there is information that would get deleted in the process: are there any rows where this field contains something other than a decimal(12,2).
I'm stumped how to do this. Apparently there isn't a string function like is_numeric() in PHP. I already tried casting the field to decimal and then comparing it with the original string, but this returns TRUE even for obvious cases where it should not:
select ('abc' = convert('abc', decimal(12,2)));
returns 1
Any help? How do I find out if a string contains something other than a decimal in MySQL? Thanks.
Stupid me, I have to cast twice (to decimal and back to char), which makes it work:
select ('abc' = convert(convert('abc', decimal(12,2)), char(255)));
returns 0
Thanks.
If you want to examine if the strings are actually floating points numbers, you could also use a regular expression. The following regex can help :)
SELECT '31.23' REGEXP '^[[:digit:]]+([.period.][[:digit:]]+)?$'; # returns 1
SELECT '31' REGEXP '^[[:digit:]]+([.period.][[:digit:]]+)?$'; # returns 1
SELECT 'hey' REGEXP '^[[:digit:]]+([.period.][[:digit:]]+)?$'; # returns 0
Related
I am trying to pull a product code from a long set of string formatted like a URL address. The pattern is always 3 letters followed by 3 or 4 numbers (ex. ???### or ???####). I have tried using REGEXP and LIKE syntax, but my results are off for both/I am not sure which operators to use.
The first select statement is close to trimming the URL to show just the code, but oftentimes will show a random string of numbers it may find in the URL string.
The second select statement is more rudimentary, but I am unsure which operators to use.
Which would be the quickest solution?
SELECT columnName, SUBSTR(columnName, LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", columnName), LENGTH(columnName) - LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", REVERSE(columnName))) AS extractedData FROM tableName
SELECT columnName FROM tableName WHERE columnName LIKE '%___###%' OR columnName LIKE '%___####%'
-- Will take a substring of this result as well
Example Data:
randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc
In this case, the desired string is "xyz123" and the location of said pattern is variable based on each entry.
EDIT
SELECT column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), SUBSTR(column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), LENGTH(column) - LOCATE(column REGEXP "^.*[a-zA-Z]{3}[0-9]{3,4}", REVERSE(column))) AS extractData From mainTable
This expression is still not grabbing the right data, but I feel like it may get me closer.
I suggest using
REGEXP_SUBSTR(column, '(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])')
Details:
(?<=[&?]random_code=[^&#]{0,256}-) - immediately on the left, there must be & or &, random_code=, and then zero to 256 chars other than & and # followed with a - char
[a-zA-Z]{3} - three ASCII letters
[0-9]{3,4} - three to four ASCII digits
(?![^&#]) - that are followed either with &, # or end of string.
See the online demo:
WITH cte AS ( SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc' val
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz4567&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz89&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz00000&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-aaaaa11111&hello_world=us&etc_etc')
SELECT REGEXP_SUBSTR(val,'(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])') output
FROM cte
Output:
I'd make use of capture groups:
(?<=[=\-\\])([a-zA-Z]{3}[\d]{3,4})(?=[&])
I assume with [^=\-] you wanted to capture string with "-","\" or "=" in front but not include those chars in the result. To do that use "positive lookbehind" (?<=.
I also added a lookahead (?= for "&".
If you'd like to fidget more with regex I recommend RegExr
I need to extract everything after the last '=' (http://www.domain.com?query=blablabla - > blablabla) but this query returns the entire strings. Where did I go wrong in here:
SELECT RIGHT(supplier_reference, CHAR_LENGTH(supplier_reference) - SUBSTRING('=', supplier_reference))
FROM ps_product
select SUBSTRING_INDEX(supplier_reference,'=',-1) from ps_product;
Please use this for further reference.
Try this (it should work if there are multiple '=' characters in the string):
SELECT RIGHT(supplier_reference, (CHARINDEX('=',REVERSE(supplier_reference),0))-1) FROM ps_product
Try this in MySQL.
right(field,((CHAR_LENGTH(field))-(InStr(field,','))))
In MySQL, this works if there are multiple '=' characters in the string
SUBSTRING(supplier_reference FROM (LOCATE('=',supplier_reference)+1))
It returns the substring after(+1) having found the the first =
If your string is
str = 'abc=def=ghi'
To select to the right:
select substring_index(str,'=',-1) from tablename ==> result is 'ghi'
select substring_index(str,'=',-2) from tablename ==> result is 'def=ghi'
To select to the left
select substring_index(str,'=',-1) from tablename ==> result is 'abc'
select substring_index(str,'=',2) from tablename ==> result is 'abc=def'
I've been working on something similar and after a few tries and fails came up with this:
Example:
STRING-TO-TEST-ON = 'ab,cd,ef,gh'
I wanted to extract everything after the last occurrence of "," (comma) from the string... resulting in "gh".
My query is:
SELECT SUBSTR('ab,cd,ef,gh' FROM (LENGTH('ab,cd,ef,gh') - (LOCATE(",",REVERSE('ab,cd,ef,gh'))-1)+1)) AS `wantedString`
Now let me try and explain what I did ...
I had to find the position of the last "," from the string and to calculate the wantedString length, using LOCATE(",",REVERSE('ab,cd,ef,gh'))-1 by reversing the initial string I actually had to find the first occurrence of the "," in the string ... which wasn't hard to do ... and then -1 to actually find the string length without the ",".
calculate the position of my wantedString by subtracting the string length I've calculated at 1st step from the initial string length:
LENGTH('ab,cd,ef,gh') - (LOCATE(",",REVERSE('ab,cd,ef,gh'))-1)+1
I have (+1) because I actually need the string position after the last "," .. and not containing the ",". Hope it makes sense.
all it remain to do is running a SUBSTR on my initial string FROM the calculated position.
I haven't tested the query on large strings so I do not know how slow it is. So if someone actually tests it on a large string I would very happy to know the results.
For SQL Management studio I used a variation of BWS' answer. This gets the data to the right of '=', or NULL if the symbol doesn't exist:
CASE WHEN (RIGHT(supplier_reference, CASE WHEN (CHARINDEX('=',supplier_reference,0)) = 0 THEN
0 ELSE CHARINDEX('=', supplier_reference) -1 END)) <> '' THEN (RIGHT(supplier_reference, CASE WHEN (CHARINDEX('=',supplier_reference,0)) = 0 THEN
0 ELSE CHARINDEX('=', supplier_reference) -1 END)) ELSE NULL END
SELECT NULLIF(SUBSTRING_INDEX(column, '=', -1), column)
Please help me resolve my query when using query - I just want to subtract a few characters and then use the % to find the matching LIKE:
select * from `providers` WHERE `name` LIKE SUBSTR('telin',1,4)%
Please let me know what i'm doing wrong, any kind of help is greatly appreciated!
Assuming telin is a column name rather than the literal string, it should be quoted in backticks. If it is the literal string, then there is obviously no need to extract a substring from it. I suspect however, that it was the result of a PHP variable you pasted here after echoing out the full query, then it is correctly single-quoted.
Anyway, you will need to concatenate the SUBSTR() result onto the '%' via CONCAT():
SELECT * FROM `providers` WHERE `name` LIKE CONCAT(SUBSTR(`telin`,1,4), '%');
But better would be to use LEFT() to compare the first 4 characters of each:
SELECT * FROM `providers` WHERE LEFT(`name`, 4) = LEFT(`telin`,4);
I have this table under user_name='high'
function_description :
akram is in a date
test
akram is studying
test4
kheith is male
test3
I want a query that returns results of field that have at least an 'akram'
SELECT *
FROM functions
WHERE 'isEnabled'=1
AND 'isPrivate'=1
AND user_name='high'
AND function_description LIKE '%akram%'
and this returns absolutely nothing!
Why?
You are listing the column names as if they are strings. This is why it returns nothing.
Try this:
SELECT *
FROM functions
WHERE user_name='high'
AND function_description LIKE '%akram%'
edit: After trying to re-read your question... are isEnabled and isPrivate columns in this table?
edit2: updated.. remove those unknown columns.
You are comparing strings 'isEnabled' with integer 1, which likely leads to the integer being converted to a string, and the comparison then fails. (The alternative is that the string is converted to an integer 0 and the comparison still fails.)
In MySQL, you use back-quotes, not single quotes, to quote column and table names:
SELECT *
FROM `functions`
WHERE `isEnabled` = 1
AND `isPrivate` = 1
AND `user_name` = 'high'
AND `function_description` LIKE '%akram%'
In standard SQL, you use double quotes to create a 'delimited identifier'; in Microsoft SQL Server, you use square brackets around the names.
Please show the schema more carefully (column names, sample values, types if need be) next time.
I know it is not an appropriate technique to have a structure of MySQL table as such, but I have to work with such. The problem is, that the field in table has value with comma seperated integers, like "1,3,5,7,10" and I want the query to return rows, in which field has a to the query passed number in it, like:
SELECT * FROM `table` WHERE '5' IN (`field_in_table`)
However, it does not work, if, in this case, '5' is not the first number in the field.
Any advises, how to solve that?
Thanks in advance,
Regards,
Jonas
Have a look at
FIND_IN_SET
Returns a value in the range of 1 to N
if the string str is in the string
list strlist consisting of N
substrings. A string list is a string
composed of substrings separated by
“,” characters. If the first argument
is a constant string and the second is
a column of type SET, the
FIND_IN_SET() function is optimized to
use bit arithmetic. Returns 0 if str
is not in strlist or if strlist is the
empty string.
You could use WHERE field_in_table LIKE '%5%' instead.
Of course, the problem would be, '1,59,98' would return as wel.
SELECT * FROM table WHERE field_in_table LIKE '%5'");
should work
You could try
SELECT *
FROM table
WHERE '%,5,%' LIKE field_in_table OR
'%,5' LIKE field_in_table OR
'5,%' LIKE field_in_table;
A better approach might be to use regular expressions, a subject on which I am not an authority.
SELECT *
FROM table
WHERE FIELD LIKE '%,5,%' OR
FIELD LIKE '5,%' OR
FIELD LIKE '%,5'